At an academically challenging high school, the average GPA of a high school senior is known to be normally distributed with a variance of 0.25. A sample of 20 seniors is taken and their average GPA is found to be 2.71. Develop a 90% confidence interval for the population mean GPA.
A) [2.5261, 2.8939]
B) [2.5667, 2.8533]
C) [2.6180, 2.8020]
D) [2.6384, 2.7816]

Question

Quizplus · Accepted Answer

The confidence interval for the population mean, when the population variance is known and the sample size is small (n < 30), can be calculated using the formula for the z-interval: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\sigma$ is the population standard deviation (the square root of the variance), $n$ is the sample size, and $z$ is the z-score corresponding to the desired confidence level. Given a variance of 0.25, the standard deviation $\sigma$ is 0.5. For a 90% confidence interval, the z-score is approximately 1.645. Plugging in the given values: $2.71 \pm 1.645 \frac{0.5}{\sqrt{20}}$ yields the interval [2.5261, 2.8939].

At an Academically Challenging High School, the Average GPA of a High